Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $a \neq 0$. $x = \dfrac{a^2 + 3a - 54}{-5a - 45} \times \dfrac{2a - 6}{5a - 30} $
First factor the quadratic. $x = \dfrac{(a - 6)(a + 9)}{-5a - 45} \times \dfrac{2a - 6}{5a - 30} $ Then factor out any other terms. $x = \dfrac{(a - 6)(a + 9)}{-5(a + 9)} \times \dfrac{2(a - 3)}{5(a - 6)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (a - 6)(a + 9) \times 2(a - 3) } { -5(a + 9) \times 5(a - 6) } $ $x = \dfrac{ 2(a - 6)(a + 9)(a - 3)}{ -25(a + 9)(a - 6)} $ Notice that $(a + 9)$ and $(a - 6)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 2\cancel{(a - 6)}(a + 9)(a - 3)}{ -25(a + 9)\cancel{(a - 6)}} $ We are dividing by $a - 6$ , so $a - 6 \neq 0$ Therefore, $a \neq 6$ $x = \dfrac{ 2\cancel{(a - 6)}\cancel{(a + 9)}(a - 3)}{ -25\cancel{(a + 9)}\cancel{(a - 6)}} $ We are dividing by $a + 9$ , so $a + 9 \neq 0$ Therefore, $a \neq -9$ $x = \dfrac{2(a - 3)}{-25} $ $x = \dfrac{-2(a - 3)}{25} ; \space a \neq 6 ; \space a \neq -9 $